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suter [353]
3 years ago
7

The probability that a single radar station will detect an enemy plane is 0.65. (a) How many such stations are required to be 98

% certain that an enemy plane flying over will be detected by at least one station?
Mathematics
1 answer:
Inga [223]3 years ago
4 0

Answer:

4 stations

Step-by-step explanation:

If we need to be at least 98% certain that an enemy plane flying over will be detected by at least one station, we must ensure that there is at most a 2% chance that no radar stations detect the plane.

The probability that a single radar station does not detect the plane is 0.35.

For n radar stations:

P=0.02=0.35^n\\n=\frac{log(0.02)}{log(0.35)}\\n=3.73

Rounding up to the next whole station, at least 4 stations are needed.

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43. Para determinar la altura de un árbol nos apoyamos en los siguientes triángulos semejantes que se forman entre el árbol y
MArishka [77]

Respuesta:

A.) BA = 12 m

Explicación paso a paso:

Usando triángulos similares:

BA / AC = DE / EC

BA = x; AC = 20; DE = 3; EC = 5

POR ESO ; TENEMOS :

x / 20 = 3/5

Multiplicar en cruz:

5 * x = 20 * 3

5 veces = 60

5x / 5 = 60/5

x = 12

3 0
3 years ago
A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are
Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

P = P_{1} + P_{2} + P_{3}

P_{1} is the probability that all three of them are 13-watt. So:

P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277

P_{2} is the probability that all three of them are 18-watt. So:

P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415

P_{3} is the probability that all three of them are 23-watt. So:

P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173

P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245

There is a 12.45% probability that one bulb of each type is selected.

3 0
3 years ago
106 = 4x + 10<br><br>What is the answer to this​
Svet_ta [14]

Answer:

24

Step-by-step explanation:

24 x 4 = 96

96 + 10 = 106

I hope this helps!

3 0
2 years ago
Read 2 more answers
The distance on a map between two locations is 2.25 inches. If = 24 miles, what the actual distance between the two locations?
almond37 [142]
The answer would be C.

2.25 + 24 = 26.25 miles
5 0
2 years ago
What is the answer to 8x - 4 = 7x - 1
mojhsa [17]
8x-7x=-1+4
1x/1 =3/1
x=3


6 0
3 years ago
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